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Pretty easy relative max problem calc 3

  1. Dec 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Locate all relative max,min,and saddle points if any.

    f(x,y)=x2+xy+y2-3x

    fx=2x+y-3

    fy=x+2y

    Skipping some algebra I get the critical points (2,-1)

    fxx=2
    fyy=2

    d=fxx*fyy-f(x0,y0)2

    d=4-9=-5

    I know I'm messing up at f(x0,y0)

    I'm simply plugging in my c.p points (2,-1) into

    f(x,y)=x2+xy+y2-3x

    but seem to keep getting -3

    While the value 1 is suppose to come out

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 5, 2012 #2

    Dick

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    It looks like you have your formula for d wrong. It's supposed to be d=fxx*fyy-(fxy)^2. fxy is the mixed second derivative.
     
  4. Dec 5, 2012 #3
    I'm a bit confused what do you mean mixed second derivative?
     
  5. Dec 5, 2012 #4

    Dick

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    First take the derivative of f with respect to x to get fx. Then take the derivative of fx with respect of y to get fxy. That's the mixed second derivative. Or you can do it in the other order and get fyx. They should be the same.
     
    Last edited: Dec 5, 2012
  6. Dec 5, 2012 #5
    Got it, thanks Dick.
     
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